Answer:![\vec{v_R}=\hat{i}[-329.11]+\hat{j}[516.18]](https://tex.z-dn.net/?f=%5Cvec%7Bv_R%7D%3D%5Chat%7Bi%7D%5B-329.11%5D%2B%5Chat%7Bj%7D%5B516.18%5D)
Explanation:
Given
Plane is initially flying with velocity of magnitude ![v=600\ mph](https://tex.z-dn.net/?f=v%3D600%5C%20mph)
at angle of
with North towards west
Velocity of plane airplane can be written as
![v_a=600(-\sin 40\hat{i}+\cos 40\hat{j})](https://tex.z-dn.net/?f=v_a%3D600%28-%5Csin%2040%5Chat%7Bi%7D%2B%5Ccos%2040%5Chat%7Bj%7D%29)
Now wind is encountered with speed of
at angle of ![N45^{\circ}E](https://tex.z-dn.net/?f=N45%5E%7B%5Ccirc%7DE)
![v_w=80(\cos 45\hat{i}+\sin 45\hat{j})](https://tex.z-dn.net/?f=v_w%3D80%28%5Ccos%2045%5Chat%7Bi%7D%2B%5Csin%2045%5Chat%7Bj%7D%29)
resultant velocity
![\vec{v_R}=\vec{v_a} +\vec{v_w}](https://tex.z-dn.net/?f=%5Cvec%7Bv_R%7D%3D%5Cvec%7Bv_a%7D%20%2B%5Cvec%7Bv_w%7D)
![\vec{v_R}=600(-\sin 40\hat{i}+\cos 40\hat{j})+ 80(\cos 45\hat{i}+\sin 45\hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Bv_R%7D%3D600%28-%5Csin%2040%5Chat%7Bi%7D%2B%5Ccos%2040%5Chat%7Bj%7D%29%2B%2080%28%5Ccos%2045%5Chat%7Bi%7D%2B%5Csin%2045%5Chat%7Bj%7D%29)
![\vec{v_R}=\hat{i}[-385.67+56.56]+\hat{j}[459.62+56.56]](https://tex.z-dn.net/?f=%5Cvec%7Bv_R%7D%3D%5Chat%7Bi%7D%5B-385.67%2B56.56%5D%2B%5Chat%7Bj%7D%5B459.62%2B56.56%5D)
![\vec{v_R}=\hat{i}[-329.11]+\hat{j}[516.18]](https://tex.z-dn.net/?f=%5Cvec%7Bv_R%7D%3D%5Chat%7Bi%7D%5B-329.11%5D%2B%5Chat%7Bj%7D%5B516.18%5D)
for direction ![\tan \theta =\frac{516.18}{329.11}](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%20%3D%5Cfrac%7B516.18%7D%7B329.11%7D)
![\tan \theta =1.568](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%20%3D1.568)
west of North
Answer:
The change in entropy ΔS = 0.0011 kJ/(kg·K)
Explanation:
The given information are;
The mass of water at 20.0°C = 1.0 kg
The mass of water at 80.0°C = 2.0 kg
The heat content per kg of each of the mass of water is given as follows;
The heat content of the mass of water at 20.0°C = h₁ = 83.92 KJ/kg
The heat content of the mass of water at 80.0°C = h₂ = 334.949 KJ/kg
Therefore, the total heat of the the two bodies = 83.92 + 2*334.949 = 753.818 kJ/kg
The heat energy of the mixture =
1 × 4200 × (T - 20) = 2 × 4200 × (80 - T)
∴ T = 60°C
The heat content, of the water at 60° = 251.154 kJ/kg
Therefore, the heat content of water in the 3 kg of the mixture = 3 × 251.154 = 753.462
The change in entropy ΔS = ΔH/T = (753.818 - 753.462)/(60 + 273.15) = 0.0011 kJ/(kg·K).
Answer:
![v=\frac{150km}{2h}=75\frac{km}{h}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B150km%7D%7B2h%7D%3D75%5Cfrac%7Bkm%7D%7Bh%7D)
Explanation:
We can use the equation for the speed
![v=\frac{x}{t}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D)
where x is the distance and t the time. In this case we know that the time spent was 2 hours and the distance was 150km. By replacing we have
![v=\frac{150km}{2h}=75\frac{km}{h}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B150km%7D%7B2h%7D%3D75%5Cfrac%7Bkm%7D%7Bh%7D)
I hope this useful for you
regards